Need help conceptualizing the constant "e"

[Fly]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]

Here is a cooler problem, imo: show that the expected value of the # of people who get their hat back is 1, independent of n.

[/ QUOTE ]

How is this cooler? This is way easier to solve than the original problem, just use <font color="white"> indicator functions </font> &lt;---- answer in white.

[/ QUOTE ]
Cooler in that it admits several elegant and simple solutions. To me, easier problems are cooler.

[/ QUOTE ]

Can't disagree with that =).

That post also jarred my memory and reminded me of an amazingly simple solution to the 1st problem using exponential generating functions. Thanks [img]/images/graemlins/smile.gif[/img]

[blah_blah]

[ QUOTE ]
[ QUOTE ]

Here is a cooler problem, imo: show that the expected value of the # of people who get their hat back is 1, independent of n.

[/ QUOTE ]

How is this cooler? This is way easier to solve than the original problem, just use <font color="white"> indicator functions </font> &lt;---- answer in white.

[/ QUOTE ]

sure, this provides an easy solution, but it's essentially a linear technique and thus isn't particularly useful if you want to generalize the problem.

what is

\sum_{\sigma\in S_n} [\fix(\sigma)]^2

[thylacine]

[ QUOTE ]

Here is a cooler problem, imo: show that the expected value of the # of people who get their hat back is 1, independent of n.

[/ QUOTE ]

Hmm, how about n(1/n)=1!
uA

[David Sklansky]

[ QUOTE ]
[ QUOTE ]

Here is a cooler problem, imo: show that the expected value of the # of people who get their hat back is 1, independent of n.

[/ QUOTE ]

Hmm, how about n(1/n)=1!
uA

[/ QUOTE ]

Obviously. And while I'm mad I just got to this thread so you posted it first, I'm quite happy that boris was nice enough to post a question that perfectly shows why clever amateur will sometimes beat not so clever pros.

Permit me to answer it in a way that everyone will understand.

There is 1000 players in a tournament redrawing for seats. The RIO is paying five thousand dollars to any player who gets his own seat. Each player has a one in a thousand chance of making a thousand dollars. Each player has an EV of $5.

I go around buying up everyone's EV for face value. It cost me five grand. Each purchase is a break even purchase for me. Thus the whole deal is a break even thing for me. Therefore my expected payoff from the Rio (which ranges from zero to 5mil) is the $5000 I paid. If the expected value of my prize is $5000, the expected number of matches is one. And of course this would work for any size tournament.

[borisp]

[ QUOTE ]
[ QUOTE ]

Here is a cooler problem, imo: show that the expected value of the # of people who get their hat back is 1, independent of n.

[/ QUOTE ]

Hmm, how about n(1/n)=1!
uA

[/ QUOTE ]
Ok, mr genius, show that the outcomes are independent. I know this is "intuitively obvious" but actually providing a proof is an altogether different matter.

And for once, I actually agree with Sklansky rigor. Tournament reseating is the perfect way to conceptualize this problem, for those who already have experience with poker tournaments.

[Subfallen]

[ QUOTE ]

And for once, I actually agree with Sklansky rigor. Tournament reseating is the perfect way to conceptualize this problem, for those who already have experience with poker tournaments.

[/ QUOTE ]

I can't even tell when you're being sarcastic anymore, lol.

What needs to be proven re: independent outcomes? Algebraically A's probability of receiving his seat doesn't change after B receives an unknown seat:
<font color="white">..</font>(1/n)(0) + ((n-1)/n)(1/(n-1)) = 1/n

So it seems to follow by induction that no B,C,D...Z...(n-1) prior assignments would ever change the algebraic probability for A receiving his seat. Is algebra not good enough here?

[borisp]

[ QUOTE ]
[ QUOTE ]

And for once, I actually agree with Sklansky rigor. Tournament reseating is the perfect way to conceptualize this problem, for those who already have experience with poker tournaments.

[/ QUOTE ]

I can't even tell when you're being sarcastic anymore, lol.

What needs to be proven re: independent outcomes? Algebraically A's probability of receiving his seat doesn't change after B receives an unknown seat:
<font color="white">..</font>(1/n)(0) + ((n-1)/n)(1/(n-1)) = 1/n

So it seems to follow by induction that no B,C,D...Z...(n-1) prior assignments would ever change the algebraic probability for A receiving his seat. Is algebra not good enough here?

[/ QUOTE ]
When something is obvious, it has not yet overcome its requirement of being written down.

However, you have provided a proof of what I was demanding. My point was that this notion (independence) must be acknowledged, otherwise the solution is incomplete.

And with regard to sarcasm, etc...if you are able to develop a method for determining this sort of thing, plz to be sending it to my home base, for we would be greatly appreciating such an algorithm [img]/images/graemlins/smile.gif[/img]

[David Sklansky]

Now please make yourself useful and go over to my thread "Improving On Buffet And Desert Cat" on the Business Forum and explain to DeserstCat that I'm right. (Even though my argument doesn't meet your brand of rigor.)

[borisp]

[ QUOTE ]
Now please make yourself useful and go over to my thread "Improving On Buffet And Desert Cat" on the Business Forum and explain to DeserstCat that I'm right. (Even though my argument doesn't meet your brand of rigor.)

[/ QUOTE ]
Like, seriously, this is a loaded request.

One way to explain your point in the business forum is to point out that history books are only written by the winners. This sufficiently discredits the opinion(s) of Buffet (or whomever) enough...or so I think...The point is that an informed opinion is worth more than the literature assigns it, since the literature has a natural bias.

In any event, I'll do my best. And if I do, I'm only doing this for irrational hero worship. I actually DID wear the cover off of my theory of poker book.

[ADDboy]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]

Here is a cooler problem, imo: show that the expected value of the # of people who get their hat back is 1, independent of n.

[/ QUOTE ]

Hmm, how about n(1/n)=1!
uA

[/ QUOTE ]
Ok, mr genius, show that the outcomes are independent. I know this is "intuitively obvious" but actually providing a proof is an altogether different matter.


[/ QUOTE ]

Why do we care about whether or not the outcomes are independent? The question is about expected values, and the expected value of the sum is the sum of the expected values regardless of whether or not the variables are independent.